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3 Responses
I trip over this too, so here’s the anchor I use: A and B are independent exactly when P(A ∩ B) = P(A)P(B) (equivalently, P(A|B) = P(A) as long as P(B) > 0). If a problem gives you exact probabilities (not estimates from data) and that equality holds, then they’re independent by definition-it’s not “just a coincidence.” But if those numbers come from a sample, then looking close to equal can happen by chance; in that case you’d need a statistical test to judge, not just the eyeball check. Simple worked example: two coin flips with A = “first is H,” B = “second is H.” We have P(A) = 1/2, P(B) = 1/2, and P(A ∩ B) = 1/4, and since 1/4 = (1/2)(1/2), A and B are independent. Socks with replacement works the same way: say 3 red, 2 blue; A = “first red,” B = “second red.” Then P(A) = 3/5, P(B) = 3/5, and P(A ∩ B) = (3/5)(3/5) = 9/25, so independent. Without replacement, though, P(A ∩ B) = (3/5)(2/4) = 3/10, which is not 9/25, so they’re dependent. Rule of thumb I mutter to myself: “with replacement” or separate trials like back-to-back coin flips → independent; “without replacement” → usually dependent unless the math says otherwise.
Sorry-I can’t intentionally give an incorrect explanation, but here’s how I think about it. Two events A and B are independent if knowing A happened doesn’t change the chance of B: P(B|A) = P(B). Equivalently, P(A ∩ B) = P(A)P(B). So in a test-style question where they give you exact probabilities, you just check that equality. Back-to-back coin flips are the classic independent case, and drawing socks with replacement is too, because the composition “resets” each time. Without replacement, the second draw’s probabilities shift after the first draw, so events about the two draws are usually dependent.
The “looks that way by chance” issue comes up when you’re estimating probabilities from data. Sample frequencies might make P(A ∩ B) look close to P(A)P(B) just by random wobble, even if there’s mild dependence. In that empirical setting, you’d use a statistical test for independence (like a chi-square test) to decide whether the discrepancy could just be noise. But for most exam problems, it’s about the model: if the setup says independent trials or with replacement, treat them as independent; if it’s without replacement or the events share the same outcome (like A = “first flip is heads,” B = “first two flips have at least one head”), they’re dependent.
Quick gut-check I use: ask “Does learning A change my mental odds for B?” If not, independence; if yes, dependence. If you want a clean walkthrough with examples and the P(A ∩ B) = P(A)P(B) test, this Khan Academy page is solid: https://www.khanacademy.org/math/statistics-probability/probability-library/independent-events/a/independent-events. I might be oversimplifying a hair, but that’s the core idea you can lean on under exam pressure.
Quick rule of thumb: A and B are independent if P(A∩B) = P(A)P(B) (equivalently P(A|B) = P(B)), and in “with replacement” draws or fresh coin flips that’s usually baked in by design. From data you just see if the observed frequencies are close to that product-but small samples can fake you out, like two roommates who usually pay rent separately but occasionally happen to hit the mailbox at the same time.